If xyz ≠ 0, is x (y + z) ≥ 0?
1) |y + z| = |y| + |z|
2) |x + y| = |x| + |y|
OA later
1) |y + z| = |y| + |z|
2) |x + y| = |x| + |y|
OA later
The powers of two are bloody impolite!!
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
IMO C coz if we take either statement then both y an z or x and y will have the same sign.. and hence we don't know what could be the sign of x(y + z).tohellandback wrote:If xyz ≠ 0, is x (y + z) ≥ 0?
1) |y + z| = |y| + |z|
2) |x + y| = |x| + |y|
OA later
could you elaborate the statement in bold?ketkoag wrote:IMO C coz if we take either statement then both y an z or x and y will have the same sign.. and hence we don't know what could be the sign of x(y + z).tohellandback wrote:If xyz ≠ 0, is x (y + z) ≥ 0?
1) |y + z| = |y| + |z|
2) |x + y| = |x| + |y|
OA later
but if we take both the statements then we know that x, y, z should have the same sign and then we know that x(y+z)>0..
hence C..
I dont think the equations imply they are both positive.life is a test wrote:1) |y + z| = |y| + |z| --> this implies that both y and z are positive.
e.g. if y=1 and z=-2 then |(+1) + (-2)| = -1 but |+1| + |-2| = 3 so y and z must both be positive for the eqn to hold. This however, still doesn't tell us about x (it couldn be +ve or -ve, all we know about z form the given info is that is is non-zero.
2) |x + y| = |x| + |y|--> for similar reasons as in 1), x and y must both be positive byt we don't know about z (it can be positive or negative)
1) and 2) together, we know that x, y and z are positive hence x(y+z) must be >0.
New here Create free account